Vectors 3 Question 14
14. Let $a, b, c$ be distinct non-negative numbers. If the vectors $a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+c \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}$ and $c \hat{\mathbf{i}}+c \hat{\mathbf{j}}+b \hat{\mathbf{k}}$ lie in a plane, then $c$ is
(1993, 1M)
(a) the arithmetic mean of $a$ and $b$
(b) the geometric mean of $a$ and $b$
(c) the harmonic mean of $a$ and $b$
(d) equal to zero
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Answer:
Correct Answer: 14. (b)
Solution:
- Since, three vectors are coplanar.
$$ \left|\begin{array}{lll} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{array}\right|=0 $$
Applying $C _1 \rightarrow C _1-C _2,\left|\begin{array}{lll}0 & a & c \ 1 & 0 & 1 \ 0 & c & b\end{array}\right|=0$
$\Rightarrow-1\left(a b-c^{2}\right)=0 \quad \Rightarrow \quad a b=c^{2}$